Advanced calculus proof question?

Question

show that:

||x -y||*||x +y|| ≤ ||x||² +||y||²

I've tried many things and no luck, any help would be greatly appreciateed. 10 pts!!!

Answer 1

I don't think I can give you the correct answer but knowing the proof to the Cauchy Schwarz Inequality will be very helpful.

Answer 2

I'm assuming here that you're dealing with real space, but only a couple changes are necessary to deal with complex spaces

||x-y||^2*||x+y||^2 = <x-y,x-y> * <x+y, x+y>

= (||x||^2 - 2<x,y> + ||y||^2)(||x||^2 + 2<x,y> + ||y||^2)

= (||x||^2 + ||y||^2)^2 - (2<x,y>)^2

<= (||x||^2 + ||y||^2)^2

Take the principal square root of both sides to get your inequality.

Answer 3

||x -y||*||x +y|| ≤ ||x||² + ||y||²

Think of ||a|| = √a² and ||a||² = a²

||x -y||*||x +y|| ≤ ||x||² +||y||²

√(x-y)² √(x+y)² ≤ x² + y²

√(x-y)√(x+y)√(x-y)√(x+y) ≤ x² + y²

√(x² - y²) √(x² - y²)  ÃƒÂ¢Ã‚‰Â¤  x² + y²

x² - y²  ÃƒÂ¢Ã‚‰Â¤  x² + y²

- y²  ÃƒÂ¢Ã‚‰Â¤  y²  yes, always